Math 241 - Chapter 13

University of South Carolina - Steven Derocher's Math241 topics covered, focused on 13.1-13.4 and any additional formulas given

(a=alpha) horizontal component of a projectile

x0+v0cos(a)t

(a=alpha) vertical component of a projectile

y0+v0sin(a)t-1/2(g)t^2

a (alpha)=

the angle formed by the projectile

the gravitational constant, g, equals

9\.8

to solve for the horizontal distance of a projectile

set the y component equal to 0, solve for t, then plug this value into the x component

(a=alpha) ideal projectile motion from point (x0,y0), r=

(x+|v0|(cosa)t)i+(y+|v0|(sina)t-(1/2)gt^2)j

v0 is the

initial velocity of the projectile

length of a smooth curve (r=x(t)i+y(t)j+z(t)k), L=

∫a→b sqrt((dx/dt)^2+(dy/dt)^2+dz/dt)^2)dt

arc length formula (simplified)

∫a→b|v(t)|dt

(T=tau) arc length parameter with base point P(t0), s(t)=

∫t0→t sqrt((x’T)^2+(y’T)^2+(z’T))dT

(T=tau) simplified arc length parameter, s(t)=

∫t0→t|v(T)|dT

unit tangent vector, T(t)=

v(t)/|v(t)|, the direction the particle is travelling in

speed on a smooth curve

ds/dt=|v(t)|

(k=kappa) curvature formula 1 (good for understanding) k=

|dT/ds|

(k=kappa, T=unit tangent vector) second curvature formula, k=

(1/|r’(t)|)(dT/dt)

principal unit normal vector, N=

(dT/dt)/(|dT/dt|), the direction T(t) is changing in

(k=kappa) third curvature formula (the one we typically use), k=

(|T(t)|X|N(t)|)/(|T(t)|)^3

radius of an osculating circle

1/k

bynormal vector, B=

TXN

TNB frame

k of a circle is…, k of a line is…

constant, 0

r(t) is orthogonal to r’(t) if

r(t) is differentiable and |r(t)| is constant

(d/dt)(u(f(t)))=f’(t)(u’(f(t))

define u as

(projectile motion) to find distance travelled by projectile,…

set j component equal to 0, solve for t (throw out any t’s that don’t make sense), plug t into r(t), solve

to find max height of projectile,…

solve for t where r’(t)=0, check that it is a max using 2nd derivative test (left side should be positive, right side should be negative), plug this t into r(t)